Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 7 / 2 / ii / Solution

Suppose an extension $H$ existed. Its degree would be $p$, and its order would be $2p(p-1)$ by the orbit-stabilizer theorem. A Sylow $p$-subgroup $P$ consequently has order $p$, and the Sylow theorems imply that its number $n_p$ divides $2(p-1)$ and is congruent to one modulo $p$. Since $2(p-1)<2p$, the only possibilities are $1$ and $p+1$. The latter would require $p+1\mid 2(p-1)$ and hence $p+1\mid 4$, impossible for $p\ge 5$. Thus $P$ is a normal subgroup of $H$.

A nonidentity element of $P$ is a $p$-cycle, so $P$ acts regularly on the $p$ points. Label them by ${\mathbb{F}}_p$ so that $P$ consists of translations. A permutation normalizing $P$ must have the form $x\mapsto ax+b$: its conjugation sends translation by one to translation by some $a\ne 0$, and its value at zero determines $b$. Therefore the normalizer of $P$ in $S_p$ has order $p(p-1)$. Normality would place $H$ inside this normalizer, contradicting $|H|=2p(p-1)$. Hence the specified dihedral permutation group has no one-point extension.